# Marsden's definition of Taylor Series

How does the following definition of Taylor polynomials:

$f(x_0 + h)= f(x_0) + f'(x_0)\cdot h + \frac{f''(x)}{2!}h^2+ ... +\frac{f^(k)(x_0)}{k!}\cdot h^k+R_k(x_0,h),$

where $R_k(x_0,h)=\int^{x_0+h}_{x_0} \frac{(x_0+h-\tau)^k}{k!}f^{k+1}(\tau) d\tau$

where I guess $\lim_{h\to 0} \frac{R_k(x_0,h}{h^k}=0$

differ from

$f(x)=f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+ \cdots +\frac {f^k(a)}{k!} (x-a)^k + R(x)$

where $R(x)$ is the corresponding error function.

I understand the intuition of the second definition and how it is derived but how does the first definition approximate the function $f$? Can you please show how to derive the definition or give an intuitive explanation in the way Tom Apostol does for the first definition:

I know a similar question is asked at Two definitions of Taylor polynomials but it isn't quite the same.

• The two definitions are the same. Are you confused that $x_0 +h$ in the first definition is $x$ in the second, and $a$ corresponds to $x_0$? Or is it something else? – davidlowryduda Apr 3 '16 at 1:19
• @mixedmath No, it's what purpose does the $h$ serve in the first definition? Why is it we multiply each term by $h^i$ in the series? – Red Apr 3 '16 at 1:36
• These are not different definitions: they say exactly the same thing in different notation. Where the first has $x_0$ the second has $a$, and the $h$ in the first formula is just $x-a$ in the second. Note $x_0+h=x_0+(x-x_0)=x$, so the first formula really is approximating $f(x)$. – symplectomorphic Apr 3 '16 at 4:54

These two definitions are the same. There is a dictionary between them. To demonstrate, we will begin with the first three terms from the second definition you give, which you find more intuitive.

So we write the degree three Taylor polynomial for $f$ at a point $a$, which is $$f(x) \approx f(a) + f'(a)(x-a) + \frac{1}{2} f''(a)(x-a)^2. \tag{1}$$

Intuitively, this approximation is very good for $x$ very near $a$, and it probably becomes a worse approximation as $x$ gets further from $a$. So let us name the difference between $x$ and $a$ as $h$, or rather $$h = x-a.$$ Then we can rewrite $(1)$ as $$f(a + h) \approx f(a) + f'(a) h + \frac{1}{2} f''(a) h^2.$$ As you can see, this is exactly the same as the first definition in your question, with the center of the expansion as $a$ (i.e. with $x_0 = a$).

It might be good to remember the two classical definitions for a derivative: $$f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x-a}$$ and $$f'(x_0) = \lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)}{h}.$$ These are the same concept, and the dictionary between them is the same as the dictionary between the two representations of Taylor polynomials in your question.

• As an aside, when I write about Taylor series I sometimes like to link to an explanation I once wrote for my students [disclaimer: link to my website] about Taylor series. Some have found it useful to clear up some ideas, and in particular give some intuition to what's going on. – davidlowryduda Apr 3 '16 at 4:39

For $h\ne 0,$ let $J$ be the open interval between $x_0$ to $x_0+h$ (regardless of whether $h>0$ or $h<0$. Suppose $$M_{k+1}=\sup_{t\in J}|f^{(k+1)}(t)|<\infty.$$ Then $$|R_k(x_0,h)|\leq |\int_{x_0}^{x_0+h}|x_0+h-t|^k M_{k+1}/k!\;dt|=$$ $$= |h|^{k+1}M_{k+1}/(k+1)!.$$ So $R_k(x_0,h)/h^k\to 0$ as $h\to 0.$ However without some restrictions on $f$, we cannot guarantee that $M_{k+1}<\infty.$

The intuition is: Assume that there is a series $S(x)=\sum_{j=0}^{\infty}a_jx^j$ and that that for some $r>0$ we have $|x|<r\implies f(x)=S(x).$ Assume that the series for $S$, and for each of its derivatives , can be differentiated term by term, for any $x\in (-r,r)$.(This means interchanging the order of two limiting processes, the infinite summation and the derivative.) Then $$f(0)=S(0)=0!a_0.$$ $$f'(0)=S'(0)=[\sum_{j=1}^{\infty}j a_j x^{j-1}]_{x=0}=1!a_1.$$ $$f''(0)=s''(0)=[\sum_{j=2}^{\infty}j(j-1)x^{j-2}]_{x=0}= 2!a_2.$$ Et cetera.

The series $S(x)=\sum_{j=0}^{\infty}f^{(j)}(0)x^j/j!$ is called the power series/ power series expansion of $f$ about $0$ / about $x=0$. It can happen to fail to be equal $f(x)$ except at $x=0.$

For example if $f(0)=0$ and $f(x)=\exp (-1/x^2)$ for $x\ne 0$ then $f^{(k)}(0)=0$ for all $k\geq 0$ so $S(x)=0$ for all $x$, but $x\ne 0\implies f(x)\ne 0.$